Problem: Find the value of $c$ so that $(x+3)$ is a factor of the polynomial $p(x)$. $p(x) = x^3 -4x^2 +cx+33$ $c=$
Solution: The following statements are equivalent: $(x+3)$ is a factor of $p(x)$ $p(x)$ is divisible by $(x+3)$ The remainder of $\dfrac{p(x)}{x+3}$ is $0$ We can use the polynomial remainder theorem to solve this problem: For a polynomial $p(x)$ and a number $a$, the remainder on division by $x-a$ is $p(a)$. According to the theorem, the remainder when $p(x)$ is divided by $(x+3)$, which can be rewritten as $(x-({-3}))$, is equal to $p({-3})$. We want this remainder to be equal to $0$. So let's set $p({-3})=0$ and solve this equation to find $c$. Let's plug ${x=-3}$ in $p( x) = x^3-4 x^2+c x+33$ and set that equal to $0$. $\begin{aligned} ({-3})^3-4({-3})^2 + c({-3}) + 33&=0 \\\\ -27-36-3c+33&=0 \\\\ -30 - 3c&=0 \\\\ -3c&=30 \\\\ c&=-10 \end{aligned}$ To conclude, $c=-10$